The resistive network shown below is connected to a D.C. source of 16 V. The power consumed by the network is 4 Watt. The value of R is :

  • Option 1)

    8\Omega

  • Option 2)

    16\Omega

  • Option 3)

    1\Omega

  • Option 4)

    6\Omega

 

Answers (1)

 

Heat developed in a resistor -

When a steady current flows through a resistance R for time t , the loss in electric potential energy appears as increased thermal energy(Heat H) of resistor and H=i^2Rt

The power devoleped = \frac{energy}{time}=i^2R=iR=\frac{V^2}{R}      (from Ohm's law)

- wherein

Unit of heat is joule (J)

Unit of power is watt (W)

 

 

P=\frac{v^{2}}{Req}

\Rightarrow Req=\frac{v^{2}}{P}=\frac{16^{2}}{4}=64\Omega

\Rightarrow 64=2R+R+4R+R

\Rightarrow 64=8R

\Rightarrow 8=R

\Rightarrow R=8\Omega


Option 1)

8\Omega

Option 2)

16\Omega

Option 3)

1\Omega

Option 4)

6\Omega

Most Viewed Questions

Preparation Products

Knockout JEE Main (Six Month Subscription)

- AI Coach Study Modules, - Unlimited Mock Tests, - Study Improvement Plan.

₹ 9999/- ₹ 8499/-
Buy Now
Knockout JEE Main (Nine Month Subscription)

- AI Coach Study Modules, - Unlimited Mock Tests, - Study Improvement Plan.

₹ 13999/- ₹ 12499/-
Buy Now
Test Series JEE Main 2024

Chapter/Subject/Full Mock Tests for JEE Main, Personalized Performance Report, Weakness Sheet, Complete Answer Key,.

₹ 7999/- ₹ 4999/-
Buy Now
JEE Main Rank Booster 2023

Booster and Kadha Video Lectures, Unlimited Full Mock Test, Adaptive Time Table, Faculty Support.

₹ 9999/- ₹ 6999/-
Buy Now
Knockout JEE Main (One Month Subscription)

Personalized AI Tutor and Adaptive Time Table, Self Study Material, Unlimited Mock Tests and Personalized Analysis Reports, Faculty Support.

₹ 7999/- ₹ 4999/-
Buy Now
Boost your Preparation for JEE Main 2021 with Personlized Coaching
 
Exams
Articles
Questions